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Q.
What amount of water is added in $40\, mL$ of $NaOH (0.1\, N )$ which is neutralized by $50 \,mL$ of $HCl (0.2 N ) ?$
Bihar CECEBihar CECE 2004
Solution:
(i) $N_{1} \,V_{1}=N_{2}\, V_{2}$
(ii) Amount of water to be added$ =$ total volume $- $ volume of $NaOH$
Given normality of $N a O H=N_{1}=0.1 \,N$
Volume of $N a O H=V_{1}=?$
Normality of $H C l N_{2}-0.2 \,N$
Volume of $H C l V_{2}=50 \,m L$
$N_{1} \,V_{1}=N_{2}\, V_{2}$
$0.1 \times V_{1}=0.2 \times 50$
$V_{1}=\frac{0.2 \times 50}{0.1}=100\, mL$
$V$ of $NaOH =40 \,mL$
Amount $H_{2} O$ to be added $=100-40=60\, m L$